Solve Trig Equation with 2 & -Π/6 Inside Brackets

[MironeDagains]

http://www5a.wolframalpha.com/Calculate/MSP/MSP238521i5b83i951f19c3000010ca05be63f0bfc0?MSPStoreType=image/gif&s=10&w=219.&h=85.

How do I solve this? I know the answers, as Wolphram Alpha has given me only the answers without any steps to how they derived those answers.
I know that sin(x)=√3/2 x= Π/3 & 2Π/3, and I know that the 2 to the left of the x means that I have to divide my answers by 2, but what about that -Π/6? What do I do with it? I have no clue where to go from here. I saw tons of youtube videos and websites describing how to solve trig equations, but they never talk about ones that have the 2 and -Π/6 inside the brackets!
What do I do?

 

[Svein]

Some observations to help you on your way:

• As you have already observed, [itex]\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2} [/itex] and [itex]\sin(\frac{2 \pi}{3})=\frac{\sqrt{3}}{2} [/itex]. Start with the first, which gives [itex] \sin(2x-\frac{\pi}{6})-\sin(\frac{\pi}{3})=0[/itex], afterwards use the other value.
• Remember that [itex]\sin(A)-\sin(B) = 2\sin\frac{A-B}{2}\cos\frac{A+B}{2} [/itex]

 

[Mentallic]

I do not agree with Svein's approach at all. That is a lot more confusing than it needs to be.

Let's start simple and ignore some of the details for the moment. The most important result that you have to realize is that if [itex]\sin{A}=\sin{B}[/itex] then A=B (not strictly true, but let's come back to that later). This means that if you solve for y in

[tex]\sin(y) = \frac{\sqrt{3}}{2}[/tex]

and arrive at the solutions [itex]y=\pi/3, 2\pi/3[/itex] then if you instead have to solve for x in

[tex]\sin(2x-\pi/3)=\frac{\sqrt{3}}{2}[/tex]

then you've just replaced y by [itex]2x-\pi/3[/itex] and you'll similarly arrive at the solutions [itex]2x-\pi/3 = \pi/3, 2\pi/3[/itex] which means you have two equations

[tex]2x-\pi/3 = \pi/3[/tex]
and
[tex]2x-\pi/3 = 2\pi/3[/tex]

which you can solve for x quite easily.

Once you've done that, you have to start thinking about the fact that the rule that A=B if [itex]\sin{A}=\sin{B}[/itex] isn't always true, because A and B can be [itex]2\pi[/itex] apart and their sines will still be equal. This will affect your answer because while you may have found all the solutions to y in the domain [itex]-\pi < y < \pi[/itex], it doesn't necessarily mean that you've found all of the solutions for x in the domain [itex]-\pi < x < \pi[/itex] because if you've found all of the solutions for y in that domain, then substituting that for [itex]2x-\pi/3[/itex] means that

[tex]-\pi < 2x-\pi/3 < \pi[/tex]

adding [itex]\pi/3[/itex] throughout

[tex]-2\pi/3 < 2x < 4\pi/3[/tex]

dividing through by 2

[tex]-\pi/3 < x < 2\pi/3[/tex]

So this tells us that we'd find all of the solutions for x in this smaller domain, and it also means that we're likely missing other solutions. What you should then do is get more values for y where [itex]\sin{y}=\sqrt{3}/2[/itex], solve for x and pick the solutions that fall in your required domain.